The Vertical Circle

 

     
  height h=2r  
  acceleration due to gravity g  
  Height x less than or equal to h   , x  
  Put any  k value  
       
  FIND    
  velocity at B so that the particle does not fall off  at A:  u=√(5gr)  
  vx=√(2gx)  
  velocity at B so that the particle  at A is zero:  u1=√(2gr)  
  velocity at A when velocity at B is u---  v=√(gr)  
  velocity at A when velocity at B is u1----v1=0  
  radius of the circle r:  
  velocity at P corresponding to u, vp=√(u2-2gx)  
  velocity at P corresponding to u1,  vp1=√(u12-2gx)  
  Angle between OP and OB  :θ in degree  
  OX  
  XP  
  Path length BP  
  Displacement BP  
  Path length/displacement  
  sinθ  
  cosθ  
  tanθ  
  Time period of revolution=T  
  threshhold Time period of revolution=T1 i.e particle shall reach the top and then free fall to ground.  
  Frequency f  :  
  x2 = vx2/u2 ;  
  y2 = vp2/u2 ;  
  x2+y2 =1  
  x  
  y  
  Let velocity at B be u2=√(5+x)gr  
  The trajectory shall be a circle of radius r2=(5+x)r/5=(1+0.2*k)r  
  velocity at A when velocity at B is u2---is  v2=√(gr2)  
 

When ground velocity is changed , the radius of the circle changes so that the apex velocity shall remain √(gr') and ground velocity is √(5gr'). When we revolve an object attached to a string, with increasing velocity, speed also increases at every point and the corresponding force is balanced by string tension till the string snaps.